Examples

Separable ODE

Separable ODE

Problem: Solve the equation $y^{\prime }=2xy$.
Solution: The given equation can be written

$$\frac{dy}{y}=2xdx.$$

Integrating both sides, obtain

$$\ln |y|=x^2+c.$$

Applying $e^{-}$ on both sides:

$$|y|=e^{x^2+c}.$$

Applying the exponent sum rule and writing $C=e^c$, we get

$$|y|=C{e}^{x^2}C>0.$$

Here $C>0$ because $e^c>0$ for any $c\in \mathbb{R}$. However, we can allow $C<0$ by removing the absolute values. So that family of functions is the same as this family:

$$y=C{e}^{x^2}C\in \mathbb{R}.$$

Note that the $C=0$ solution, $y=0$, is valid but isn’t generated by separation of variables because of the division by $y$ that was performed in the first step.

Integral curves

Integral curves

Problem: Solve the separable ODE:

$$\frac{dy}{dx}=\frac{x^2}{1-y^2}.$$

Solution: Rewrite this as:

$$(1-y^2)dy=x^{2}dx$$

and integrate both sides to obtain:

$$y-\frac{y^3}{3}=\frac{x^3}{3}+c.$$

This equation specifies a family of integral curves. For the integral curves, we do not need to solve for $y$ in terms of $x$. Here is a depiction:

First-order linear

First-order linear

Problem: Solve the equation:

$$x^2{y}^{\prime }+xy=x^3.$$

Solution: First divide by $x^2$ to recover the standard format:

$$y^{\prime }+x^{-1}y=x.$$

Now the integrating factor we need is $e^{\int x^{-1}dx}=e^{\ln |x|}=|x|$. Multiply everything by this factor:

$$|x|y^{\prime }+\frac{|x|}{x}y=x|x|.$$

If $x>0$ everywhere, we can drop the absolute value signs obtaining $x{y}^{\prime }+y=x^2$. If $x<0$ everywhere, we would need to introduce minus signs (recall that $|x|=-x$ and $|x|/x=-1$ when $x<0$), obtaining $-x{y}^{\prime }-y=-x^2$; but this we can multiply by $-1$ and obtain the same equation, namely:

$$x{y}^{\prime }+y=x^2.$$

Let us solve this and come back later to consider the possibility of $x$ changing signs.

The sum on the left is the output of the product rule: ${\left(xy\right)}^{\prime }=x{y}^{\prime }+y$. Now integrate in $x$ to find a family of solutions:

$$\begin{array}{rl}xy& =\int x^{2}dx=\frac{1}{3}{x}^3+c,\\ y& =\frac{1}{3}{x}^2+\frac{c}{x}.\end{array}$$

This family, for various values of $c\in \mathbb{R}$, is our proposed solution to the ODE.

Now reconsider the absolute values. Suppose $c>0$. If we take a solution curve with $x>0$, then the entire curve will satisfy $x>0$ because $y\to +\infty$ as $x\to 0^{+}$. Similarly, if we start with $x<0$, then $y\to -\infty$ as $x\to 0^{-}$. No solution crosses over $x=0$. Now if $c<0$ we can say the same thing but with $y\to -\infty$ and $y\to +\infty$, respectively, instead. Therefore our analysis using cases $x>0$ and $x<0$ is justified.

Solving an exact equation

Solving an exact equation

Problem: Solve the ODE:

$$e^{y}dx+(x{e}^y+2y)dy=0.$$

Solution: First we check that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$:

$$\begin{array}{rl}\frac{\partial M}{\partial y}& =e^y\\ \frac{\partial N}{\partial x}& =e^y+0=e^y.\end{array}$$

Now we seek $f(x,y)$ with $\frac{\partial f}{\partial x}=M=e^y$ and $\frac{\partial f}{\partial y}=N=(x{e}^y+2y)$. First integrate $e^y$ in $x$:

$$f(x,y)=\int e^{y}dx=x{e}^y+g(y)$$

where the “constant” can be any function of $y$. Now $\frac{\partial f}{\partial y}=x{e}^y+g^{\prime }(y)=x{e}^y+2y$, so $g^{\prime }(y)=2y$ and $g(y)=y^2+c$. Therefore our solution family of integral curves is given by:

$$x{e}^y+y^2=c$$

for $c\in \mathbb{R}$.